# Use Fubini's Theorem to verify that $E(Y_n)=0$

Suppose $$X_1,X_2$$ are iid random variables with common N(0,1) distribution. Define $$Y_n = \frac{X_1}{\frac{1}{n}+|X_2|}$$. Use Fubini's Theorem to verify that $$E(Y_n)=0$$ Note that as as $$n \rightarrow \infty$$ $$Y_n \rightarrow Y:= \frac{X_1}{|X_2|}$$ and that the expectation of $$Y$$ does not exists.

## Proof:

$$E(Y_n)= \int_{\Omega_1 \times \Omega_2} Y_n dP=\int_{\Omega_1}\big(\int_{\Omega_2} \frac{X_{1}(\omega_1)}{\frac{1}{n}+|X_{2}(\omega_2)|} dP_{2}(\omega_2)\big)dP_1(\omega_1)=\int_{\Omega_1}X_{1}(\omega_1)\big(\int_{\Omega_2} \frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} dP_{2}(\omega_2)\big)dP_1(\omega_1)=\int_{\Omega_2} \frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} \big(\int_{\Omega_1}X_{1}(\omega_1)\big(dP_1(\omega_1) \big)dP_{2}(\omega_2)=\int_{\Omega_2} \frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} E(X_1)dP_{2}(\omega_2)=\int_{\Omega_2} 0 *\frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} dP_{2}(\omega_2)=\int_{\Omega_2} 0 dP_{2}(\omega_2)=0$$

You're nearly there. Try putting the integral over $\Omega_1$ on the inside:
• why is $X_2(\omega)$ shouldn't it be $X_2(\omega_2)$ @MarcusM Commented Oct 22, 2015 at 15:35
• @RobJames, Oops! You're right, it should be $X_2(\omega_2)$. Commented Oct 22, 2015 at 15:36
• What are $P_1$ and $P_2$? Commented Oct 22, 2015 at 15:54
• $P=P_1 \times P_2$ They are the probabilities on $\Omega_1$ and $\Omega_2$ respectively @JohnDawkins Commented Oct 22, 2015 at 16:10